FR35 · MRI

MRI Frequency Encoding: Turning Position into Pitch

One coil hears every spin at once. How does a field gradient tell it where each one sits? A magnetic field that grows across the scanner makes protons precess faster on one side than the other, so each position gets its own frequency. The coil still records one summed signal, but that signal is the Fourier transform of where the water is, and a Fourier sum turns it back into a profile. How finely the signal is sampled sets the field of view; how long it is sampled sets the pixel size.

Two water tubes sit in a magnetic field gradient. Each precesses at its own frequency, the coil records their sum, and a Fourier sum turns that record into a profile.

Frequency offset of the selected tube $-9.58\,\mathrm{kHz}$

Field of view $235\,\mathrm{mm}$

Pixel width $1.83\,\mathrm{mm}$

Where the image puts the selected tube $-45\,\mathrm{mm}$

Choose the gradient and the sampling

Readout gradient

5

G\;(\mathrm{mT/m})

Receiver bandwidth

50

1/\Delta t\;(\mathrm{kHz})

Samples per readout

128

N

Transverse decay time

30

T_2^{*}\;(\mathrm{ms})

Gradient actually played (the scanner assumes positive)

Drag a water tube along the gradient; the record and the image are acquired again as it moves.

Receiver noise

Receiver noise relative to the total signal

0.005

Seed of the receiver noise

3501

Model, units and sources

$s_n=e^{-t_n/T_2^{*}}\sum_j\rho_jw_j\,\mathrm{sinc}(k_nw_j)\,e^{-i2\pi k_nx_j}+\varepsilon_n,\quad m(x_p)=\Delta k\sum_n s_n\,e^{+i2\pi k_nx_p}$

Each tube is a uniform block of water, so its transform is exact; nothing is rasterized before sampling. The echo falls in the middle of the readout, 1 ms after the end of the 90-degree pulse plus half the readout. The time-domain check integrates the Bloch equations for 400 isochromats per tube and agrees with the Fourier formula to better than one part in a thousand.

MIT 6.3000 MRI handout (2025), pp. 11–13 · Bloch (1946), Eqs. 38a–c

How a gradient turns position into frequency

Model assumptions

One-dimensional model in the frame rotating at the Larmor frequency of the main field. Water tubes are uniform blocks of spins; the coil is uniformly sensitive. After an ideal 90-degree pulse a prephasing gradient and a readout gradient play, and the recorded samples are the analytic Fourier transform of the blocks with a single-exponential T2* decay and seeded complex Gaussian noise. The image is the inverse discrete Fourier sum of those recorded samples. No chemical shift, no gradient non-linearity, no eddy currents and no body; the numbers are a teaching model, not scanner performance.

What the coil recorded

The profile computed from that record

More numbers and how they are checked

Time between samples: $0$
Spatial-frequency step: $0$
Largest spatial frequency reached: $0$
Readout duration: $0$
Echo time: $0$
Frequency spread across the field of view: $0$
Selected tube, true centre: $0$
Selected tube, centre folded into the field of view: $0$
Transverse signal left at the end of the readout: $0$
Receiver noise relative to the total signal: $0$
Image energy over sample energy (Parseval, should be 1): $0$

The yellow profile is computed only from the recorded samples. The dashed blue outline is the model object and never passes through the reconstruction, so their difference is the honest cost of finite sampling, decay and noise.

Start with the essentials

Focus question: One coil hears every spin at once. How does a field gradient tell it where each one sits?
One-sentence intuition: A gradient adds a field $Gx$ , so a spin at $x$ runs $\bar\gamma Gx$ hertz faster than the reference. After time $t$ it has turned by $2\pi kx$ with $k=\bar\gamma Gt$ . The coil signal is then the Fourier transform of the density, sampled at those values of $k$ , and an inverse Fourier sum gives the profile back.

Core mathematical model

Signal of the whole sample

s(t)=e^{-t/T_2^{*}}\int\rho(x)\,e^{-i2\pi k(t)x}\,\mathrm{d}x,\qquad k(t)=\bar\gamma\int_0^t G(\tau)\,\mathrm{d}\tau

Rotating frame, $\bar\gamma=\gamma/2\pi=42.576\,\mathrm{MHz/T}$ for protons. The minus sign follows the MIT 6.3000 handout: a positive gradient makes the phase decrease.

Sampling sets the field of view and the pixel

\Delta k=\bar\gamma G\,\Delta t,\qquad \mathrm{FOV}=\frac{1}{\Delta k},\qquad \Delta x=\frac{\mathrm{FOV}}{N}

Samples spaced by $\Delta t$ are spaced by $\Delta k$ in spatial frequency. Anything outside one field of view folds back in.

Common difficulties

The gradient does not move the spins

Typical misconceptionThe gradient pushes water to one side, and the coil then senses where it went.

Better mental modelNothing moves. The gradient only changes how fast each spin precesses, so their phases fan out in a pattern that depends on position. The coil measures how well that fan lines up with each spatial frequency.

Run the experiment

01
Watch the arrows fan out
Start with Two tubes, two pitches and watch the stage for one cycle.
What to observe: During prephasing the arrows spread; at the echo they line up again and the coil arrow is longest; then they fan out the other way.
02
Drag a tube
Drag the yellow-outlined tube across the gradient.
What to observe: Its frequency offset changes sign at the centre, and its bump in the profile follows it until it leaves the field of view and reappears on the other side.
03
Halve the bandwidth
Choose Slow sampling folds the image.
What to observe: The field of view shrinks to 117 mm and the right-hand tube appears on the left, at its true position minus one field of view.

MRI Frequency Encoding: Turning Position into Pitch

Position written as frequency

Core mathematical model

Signal of the whole sample

Sampling sets the field of view and the pixel

Common difficulties

The gradient does not move the spins

Run the experiment

Watch the arrows fan out

Drag a tube

Halve the bandwidth

Related experiments and articles

From precessing water to k-space, undersampling and diffusion